coi lại cái phân số thứ 2

khẳng định là sai còn nếu đúng thì quy đồng lên làm đảm bảo số lớn khủng

\(\frac{x+5}{95}+\frac{x+10}{90}+\frac{x+15}{85}+\frac{x+20}{80}=-4\)

<=>  \(\frac{x+5}{95}+1+\frac{x+10}{90}+1+\frac{x+15}{85}+1+\frac{x+20}{80}+1=0\)

<=>  \(\frac{x+100}{95}+\frac{x+100}{90}+\frac{x+100}{85}+\frac{x+100}{80}=0\)

<=>  \(\left(x+100\right)\left(\frac{1}{95}+\frac{1}{90}+\frac{1}{85}+\frac{1}{80}\right)=0\)

<=>   \(x+100=0\)   (do 1/95 + 1/90 + 1/85 + 1/80 khác 0)

<=>   \(x=-100\)

Vậy...

Đáp án : B

Đáp án: B

\(a.x\in B\left(8\right)=\left\{0;8;16;24;32;...\right\}\)

\(\text{Mà }x\le30\Rightarrow x\in\left\{0;8;16;24\right\}\)

\(b.x\in B\left(15\right)=\left\{0;15;30;45;60;75;90;105;...\right\}\)

\(\text{Mà }15\le x\le90\Rightarrow x\in\left\{15;30;45;60;75;90\right\}\)

\(c.x\in B\left(5\right)=\left\{0;5;10;15;20;30;...;85;90;95;100;105;...\right\}\)

\(\text{Mà }x\le100\Rightarrow x\in\left\{0;5;10;15;20;...;95;100\right\}\)

\(\text{d. x }⋮12\Rightarrow x\in B\left(12\right)=\left\{0;12;24;36;48;60;72;84;96;...\right\}\)

\(\text{Mà }24\le x\le80\Rightarrow x\in\left\{24;48;60;72\right\}\)

Thanks bạn nha

\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

*Note: Tích của một số với 0 luôn bằng 0.

a) \(\left(1763+7634+23\cdot764\right)\cdot\left(346\cdot45-173\cdot90\right)\)

\(=\left(1763+7634+23\cdot764\right)\cdot\left(346\cdot45-173\cdot45\cdot2\right)\)

\(=\left(1763+7634+23\cdot764\right)\cdot\left(346\cdot45-346\cdot45\right)\)

\(=\left(1763+7634+23\cdot764\right)\cdot0\)

\(=0\)

b) \(N=10\cdot12\cdot14\cdot16\cdot\left(88\cdot17-44\cdot34\right)\)

\(N=10\cdot12\cdot14\cdot16\cdot\left(88\cdot17-44\cdot17\cdot2\right)\)

\(N=10\cdot12\cdot14\cdot16\cdot\left(88\cdot17-88\cdot17\right)\)

\(N=10\cdot12\cdot14\cdot16\cdot0\)

\(N=0\)

c) \(\left(80\cdot96+40\cdot1808\right):1000\)

\(=\left(80\cdot96+40\cdot904\cdot2\right):1000\)

\(=\left(80\cdot96+80\cdot904\right):1000\)

\(=\left[80\cdot\left(96+904\right)\right]:1000\)

\(=80\cdot1000:1000\)

\(=80\)